# Math4201 Topology I (Lecture 30)

## Compactness

### Compactness in Metric Spaces

#### Limit point compactness

A topological space $X$ is limit point compact if every infinite subset of $X$ has a limit point in $X$.

- Every compact space is limit point compact.

#### Sequentially compact

A topological space $X$ is sequentially compact if every sequence in $X$ has a convergent subsequence.

#### Theorem of equivalence of compactness in metrizable spaces

If $(X,d)$ is a metric space then the following are equivalent:

1. $X$ is compact.
2. $X$ is limit point compact.
3. $X$ is sequentially compact.

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<summary>Proof</summary>

(1) $\implies$ (2):

We proceed by contradiction,

Let $X$ be compact and $A\subseteq X$ be an infinite subset of $X$ that doesn't have any limit points.

Then $X-A$ is open because any $x\in X-A$ isn't in the closure of $A$ otherwise it would be a limit point for $A$, and hence $x$ has an open neighborhood contained in the complement of $A$.

Next, let $x\in A$. Since $x$ isn't a limit point of $A$, there is an open neighborhood $U_x$ of $x$ in $X$ that $U_x\cap A=\{x\}$. Now consider the open covering of $X$ given as

$$
\{X-A\}\cup \{U_x:x\in A\}
$$

This is an open cover because either $x\in X-A$ or $x\in A$ and in the latter case, $x\in U_x$ since $X$ is compact, this should have a finite subcover. Any such subcover should contain $U_x$ for any $x$ because $U_x$ is the element in the subcover for $x$.

This implies that our finite cover contains infinite open sets, which is a contradiction.

---

(2) $\implies$ (3):

Let $\{x_n\}_{n\in\mathbb{N}}$ be an arbitrary sequence in $X$.

Since $d(z,x_{n_k})\leq \frac{1}{k}$ the subsequence $(x_{n_k})$ converges to $z$.

This completes the proof.

except possibly $z$.

Now we consider

$$
B_{r_k}(z)\text { with } r_k=\min \left(\frac{1}{k}, d_k\right)
$$

This ball has a point $x_{n_k}$ from $\{x_n\}$ which isn't equal to $z$.

$r_k\leq d_k\implies n_k\geq n_{k-1}$.

Since $z$ is a limit point of $\{x_n\}$, there exists $x_{n_k}$ such that $d(z,x_{n_k})<\frac{1}{k}$. So $x_{n_k}\in B_{r_k}(z)$.

So, we have a convergent subsequence $(x_{n_k})$.

---

(3) $\implies$ (1):

First wee prove the analogue of Lebesgue number lemma for a sequentially compact space $(X,d)$.

Let $\{U_\alpha\}_{\alpha\in I}$ be an open covering of $X$. **By contradiction**, assume that for any $\delta>0$, there are two points $x,x'$ with $d(x,x')<\delta$ don't belong to the same open set in the covering.

Take $\delta=\frac{1}{n}$, and let $x_n,x_n'$ be the points as above, then $d(x_n,x_n')<\frac{1}{n}$.

$x_n,x_n'$ don't belong to the same open set in $\{U_\alpha\}_{\alpha\in I}$.

By assumption $\{x_n\}$ is convergent after passing to a subsequence

$$
\{x_{n_k}\}_i
$$

Let $y$ be the limit of this subsequence and $U_\alpha$ be an element of the open covering containing $y$. There is $\epsilon>0$ such that $B_\epsilon(y)\subseteq U_\alpha$.

If $k$ is large enough, then $x_{n_k}\in B_{\epsilon/2}(y)$ and $d(x_{n_k},x_{n_k}')<\epsilon/2$. (take $k$ such that $\frac{1}{n_k}<\epsilon/2$)

Then $d(x_{n_k}',y)<\epsilon/2$ this implies that $x_{n_k}'\in U_\alpha$.

Thus, $d(x_{n_k}',y)\leq d(x_{n_k}',x_{n_k})+d(x_{n_k},y)<\epsilon/2+\epsilon/2=\epsilon$.

So, $x_{n_k}'\in B_\epsilon(y)\subseteq U_\alpha$.

This is a contradiction.

Next we show that for any $\epsilon$, there are

$$
y_1,y_2,\cdots,y_k
$$

such that $X=\bigcup_{i=1}^k B_{\epsilon}(y_i)$.

Let's assume that it's not true and construct a sequence of points inductively in the following way:

- Pick $y_1$ be arbitrary point in $X$.
- In the $k$-th step, if $X\neq B_{\epsilon}(y_1)\cup \cdots \cup B_{\epsilon}(y_k)$, then pick $y_{k+1}\notin B_{\epsilon}(y_1)\cup \cdots \cup B_{\epsilon}(y_k)$.
- In particular, $d(y_{k+1},y_j)\geq \epsilon$ for all $j<k$.

By iteration, this process we obtain a sequence such such that the distance between any two elements is at most $\epsilon$.

This sequence cannot have a converging subsequence which is a contradiction.

To prove the compactness of $X$, take an open covering $\{U_\alpha\}_{\alpha\in I}$ of $X$ and let be $\delta>0$ such that any set with diameter at least $\delta$ is one of the $U_\alpha$'s. Let also $y_1,y_2,\cdots,y_k\in X$ be chosen such that $B_{\frac{\delta}{2}}(y_1)\cup \cdots \cup B_{\frac{\delta}{2}}(y_k)=X$.

Since diameter of $B_{\frac{\delta}{2}}(y_i)$ is less than $\delta$, it belongs to $U_\alpha$ for some $\alpha\in I$.

Then $\{U_{\alpha_i}\}_{i=1}^k$ is a finite subcover of $X$.

This completes the proof.

</details>